Triangles The following examples explore aspects of triangles:

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1 Tringles The following exmples explore spects of tringles:

2 xmple 1: ltitude of right ngled tringle +

3 xmple : tringle ltitude of the symmetricl ltitude of n isosceles x x - 4 +x

4 xmple 3: ltitude of the non-symmetricl ltitude of n isosceles tringle x x x x

5 xmple 4: ltitude of tringle defined y side lengths Heron s formul sys tht the re of tringle is ( + + c)( + c)( + c)( + + c) 4 You ll need to expnd the expression to see the reltionship to the ltitude elow: c 4 c c 4 c 4 c

6 xmple 5: ltitude of n Isosceles Tringle n isosceles tringle is defined y se length nd one ngle. Wht is its ltitude? sin(θ) cos(θ) θ

7 xmple 6: Tringle res Two sides nd the included ngle: θ sin(θ)

8 Two sides nd the non included ngle: - sin(θ) + cos(θ) sin(θ) θ

9 Three sides: ++c +-c -+c -++c 4 c

10 xmple 7: Intersection of the Medins of Right ngled Tringle Medins will e esy to specify when we hve the Midpoint construction. However, they cn lso e specified simply y setting the distnce to the end of the medin to e hlf the specified length of the side: F + 3

11 xmple 8: entroid of generl tringle We look t the centroid (intersection of the medins). We check the length of the medin nd the loction of the incenter on tht medin: c 9 F + - c 4 c c We see tht the centroid is locted /3 of the wy down the medin. lterntively, if we constrin the tringle y specifying its vertices, we otin this formul for the loction of the centroid:

12 x 1,y 1 F x 0 +x 1 +x y 0 +y 1 +y, 3 3 x 0,y 0 x,y

13 xmple 9: ngle of Medin Given tringle with sides length, nd included ngle θ we derive the ngle the medin mkes with the other side of the tringle: rctn - sin(θ) -+ cos(θ) rctn - sin(θ) - + θ

14 xmple 10: Orthocenter We look t the orthocenter (intersection of the ltitudes) of tringle. gin we derive n expression for its distnce from vertex of the tringle. F - +c ++c (+-c) (-+c) (-++c) c

15 xmple 11: Orthocenter oordintes This exmple is tken from the pper Lernen mit einem neuen Tschenrechner Given tht nd re locted t points (,) nd (c,) respectively, nd tht point cn move long the line y=d, we find the coordintes of the intersection of the ltitudes. The locus is qudrtic, s shown oth y its shpe nd y its eqution. -X - - c+ d+x (+c)+y (-d)=0 (t,d) F (,) (c,)

16 xmple 1: ircumcircle rdius The center of the circumcircle of tringle is the intersection of the perpendiculr isectors. xmining the expression for the distnce from one vertex to the intersection point shows tht it is symmetric. Hence will e the sme, whichever vertex is chosen (try it). F c ++c -(+-c) (-+c) (--c) c

17 Of course, more direct pproch is just to drw the circumcircle: c ++c (+-c) (-+c) (-++c) c

18 xmple 13: ircumcircle rdius in terms of vectors If we hve vectors defining circle, wht is the circle s rdius: u 0 -u 1 u 0 v + 0 -u 0 u 1 v 0 -u 0 v 1 -u 0 +u 1 + -v 0 +v 1 + u 0 v + 0 -v +v +v u 1 v 0 -u 0 v 1 -u 0 +u 1 + -v 0 +v 1 u 1 u 0 v 1 v 0

19 xmple 14: Incircle Rdius +-c -+c -++c ++c c

20 xmple 15: Vector joining the incircle with the vertex of tringle defined y vectors u 1 u 0 +v0 +u0 u 1 +v1 u 0 v 0 u 0 +v0 + u1 +v1 + u0 -u 1 + -v0 +v 1 v 1 u 0 +v0 +v0 u 1 +v1 F u 0 +v0 + u1 +v1 + u0 -u 1 + -v0 +v 1 u 1 v 1 xmine the expression the denomintor of ech coefficient is the perimeter of the tringle, so the vector cn e written: u v 0 0 u u u u P 1 P + v u 1 0 0

21 xmple 16: Incircle enter in rycentric oordintes x 1,y 1 x,y x 0,y 0 x -x 0 +x 1 + y 0 -y 1 +x 1 -x 0 +x + y 0 -y +x 0 -x 1 +x + y 1 -y -x 0 +x 1 + y 0 -y 1 + -x 0 +x + y 0 -y + -x 1 +x + y 1 -y y -x 0 +x 1 + y 0 -y 1 +y 1 -x 0 +x + y 0 -y +y 0 -x 1 +x + y 1 -y, -x 0 +x 1 + y 0 -y 1 + -x 0 +x + y 0 -y + -x 1 +x + y 1 -y If we let =, = nd c= nd let, nd e the position vectors of the points,, then the incircle center is: + + c + + c

22 xmple 17: split line How does the point of contct with the incircle +-c G H --c F c

23 xmple 18: Length of line joining vertex with the point of contct etween the incircle nd the opposite side c+ c+3 c + c - c 3 H c

24 xmple 19: Pythgors-Like igrm rw tringle (not necessrily right ngled), nd set the side lengths to e, nd c. Now drw squres on ech side, constrining them with right ngles nd lengths of sides. Now look t the distnce etween neighoring corners of the squres: - + c c c c c + -c

25 xmple 0: Penequilterl Tringle Strting with tringle whose sides re length,,c, we construct squres on ech side, join the corners of the squres, then join the midpoints of these lines to crete tringle: F J G c L K I H

26 The tringle looks to the nked eye s if it is equilterl. Try drgging the originl points nd oserve tht the new tringle still looks equilterl: I c H G K L J F Is it in fct?

27 + + 3 c c +-c -+c -++c F J c + ++c +-c -+c -++c I G c L K K I H c c +-c -+c -++c 16 Not quite, we oserve the sides re gurnteed to e close in size, ut not identicl unless the originl tringle is isosceles. In fct, the difference in squres of the sides of the new tringle is c. 4 4 Notice we cn repet the process drwing squres on JK, KL nd JL. The difference in squres of c the ensuing sides will e y repeting this process, we cn crete tringle s close s we like to n equilterl tringle, ut still not exctly one.

28 O V M F J P G N L K U I H T Q R

29 xmple 1: nother Penequilterl Tringle We cn do similr construction sed on equilterl tringles drwn on the sides of n originl tringle: Side Length c c +-c -+c -++c 8 G H re c c +5 ++c +-c -+c -++c 3 I We see tht the difference in squres of the sides of the new tringle corresponding to nd is c. If we repet the process, this tringle will eventully ecome equilterl, ut not s quickly s the previous tringle. In fct, if S(n) is the sum of the sides t the nth stge in the itertion, nd (n) is the re of the tringle on the nth stge of the itertion, the ove equtions for side length nd re show tht the following re true: 5 S( n) = S( n 1) + 8 ( n) = 3 6 ( n 1) 3 5 S( n 1) + ( n 1) 3 8 Which reltion we cn solve to give:

30 S(0) 5 (0) 8 ( S( n) ( n) ) = n

31 xmple : Folding right ngled tringle If you crete second right ngled tringle y cutting perpendiculr to the hypotenuse, then tht tringle is similr to the originl one. Here re its side lengths: x + x x

32 xmple 3: Interior point of n quilterl Tringle If point is interior to n equilterl tringle, the sum of the perpendiculr distnces to the sides of the tringle is independent of the loction of the point. The following construction shows this (try dding the three distnces). F x- y 6 G y 3 (- x+y) 3 3 x- y 3 x

33 Here is simpler wy to define this one, using perpendiculr distnce constrints. In this cse we specify x to e the perpendiculr distnce etween nd, nd y to e the perpendiculr distnce from to. We then sk for the perpendiculr distnce from to. x y - 3 +x+y Notice tht the sum of these 3 distnces is 3, which is independent of x nd y.

34 xmple 4: Right ngled tringle in semi-circle This digrm shows tht the center of the hypotenuse of right ngled tringle is hlf hypotenuse wy from the opposite vertex: d d d

35 xmple 5: Line Joining the pex to se of n isosceles tringle Tke n isosceles tringle whose equl sides hve length nd divide its se in lengths x, y. Join the dividing point to the opposite vertex, nd exmine its length. The expression is surprisingly simple: -x y x y From this result, you cn deduce, for exmple, tht for ny two chords through prticulr point interior to circle, the product of the lengths of the chord segments on either side of the point is the sme.

36 xmple 6: re of the tringle is: re of Tringle defined y vectors u 1 v 0 u 0 v - 1 u 0 v 0 u 1 v 1

37 xmple 7: Similr Tringle Lengths on similr tringle go y rtio: c c d c d c

38 xmple 8: res of tringles ounded y evins The rtios of res in this digrm re simple qudrtics in t. n you derive these qudrtics from the sine rule? re of z 4 ++c +-c -+c -++c 4 re of F z 3 t ++c +-c -+c -++c 4 - t ++c +-c -+c -++c 4 t<1 z 3 t-t re of F/ = z 4 t<1 t t F c c t re of F z 0 ++c +-c -+c -++c re of F/ = 1-3 t+3 t z 0 z ++c +-c -+c -++c c +-c -+c -++c 4 3 t ++c +-c -+c -++c t ++c +-c -+c -++c t ++c +-c -+c -++c t ++c +-c -+c -++c > t ++c +-c -+c -++c t ++c +-c -+c -++c >0 4 4

39 xmple 9: ecomposing vector into components prllel nd perpendiculr to second vector v 0 -u 1 v 0 +u 0 v 1 u 1 v 1 -u 0 -v0 u 0 u 1 v 0 -u 0 v 1 -u 0 -v0 u 0 -u 0 u 1 -v 0 v 1 u 0 v 0 -u 0 -v0 v 0 -u 0 u 1 -v 0 v 1 -u 0 -v0

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